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Why do compressed liquids heat up when allowed to expand unlike gases?
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$begingroup$
This is an experimental observation on a chromatographic column. I am trying to reconcile two observations, one with gases and one with liquids. Assume a narrow tube packed with very small 2-3 micron particles made of silica.
When we pump solvents such as methanol/water (from the left, inlet) at high flow rates, say 5 mL/min, the back-pressure is very high ~ 500 bar. The right end of the column heats up. One can feel that outlet fluid is quite warm. The outlet of the column is at atmospheric pressure. The inlet to outlet temperature difference can be as high as 10 C. In the literature it is wrongly called frictional heating as if the fluid has warmed up by rubbing against very small silica particles. I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles.
Now, gases can also be used as a mobile phase in the same packed column, e.g. pressurized carbon dioxide. This time the right end of the column is almost freezing, as expected. The pressurized gases are expanding and the gases cool down.
The question is why do compressed liquids warm up when they are allowed to expand, unlike gases? Thanks.
thermodynamics thermal-radiation
$endgroup$
add a comment |
$begingroup$
This is an experimental observation on a chromatographic column. I am trying to reconcile two observations, one with gases and one with liquids. Assume a narrow tube packed with very small 2-3 micron particles made of silica.
When we pump solvents such as methanol/water (from the left, inlet) at high flow rates, say 5 mL/min, the back-pressure is very high ~ 500 bar. The right end of the column heats up. One can feel that outlet fluid is quite warm. The outlet of the column is at atmospheric pressure. The inlet to outlet temperature difference can be as high as 10 C. In the literature it is wrongly called frictional heating as if the fluid has warmed up by rubbing against very small silica particles. I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles.
Now, gases can also be used as a mobile phase in the same packed column, e.g. pressurized carbon dioxide. This time the right end of the column is almost freezing, as expected. The pressurized gases are expanding and the gases cool down.
The question is why do compressed liquids warm up when they are allowed to expand, unlike gases? Thanks.
thermodynamics thermal-radiation
$endgroup$
$begingroup$
Are you familiar with the property of a fluid called "viscosity?"
$endgroup$
– Chet Miller
6 hours ago
1
$begingroup$
I am glad you didn't start with "Elementary, my dear Watson!"
$endgroup$
– M. Farooq
6 hours ago
$begingroup$
I don’t understand. What are you saying?
$endgroup$
– Chet Miller
6 hours ago
1
$begingroup$
I was intrigued by the way you asked if I were familiar with the fluid property called "viscosity". Of course, everyone is. If your point is viscous heating, this is exactly what that expert was refuting. He said that the heating of the liquid has nothing to do with friction of the liquid with the particles. P.S. "Elementary, my dear Watson" is a Sherlock Holmes' dialogue to his friend Dr. Watson, when he solved a difficult case and wanted to explain it to him.
$endgroup$
– M. Farooq
5 hours ago
add a comment |
$begingroup$
This is an experimental observation on a chromatographic column. I am trying to reconcile two observations, one with gases and one with liquids. Assume a narrow tube packed with very small 2-3 micron particles made of silica.
When we pump solvents such as methanol/water (from the left, inlet) at high flow rates, say 5 mL/min, the back-pressure is very high ~ 500 bar. The right end of the column heats up. One can feel that outlet fluid is quite warm. The outlet of the column is at atmospheric pressure. The inlet to outlet temperature difference can be as high as 10 C. In the literature it is wrongly called frictional heating as if the fluid has warmed up by rubbing against very small silica particles. I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles.
Now, gases can also be used as a mobile phase in the same packed column, e.g. pressurized carbon dioxide. This time the right end of the column is almost freezing, as expected. The pressurized gases are expanding and the gases cool down.
The question is why do compressed liquids warm up when they are allowed to expand, unlike gases? Thanks.
thermodynamics thermal-radiation
$endgroup$
This is an experimental observation on a chromatographic column. I am trying to reconcile two observations, one with gases and one with liquids. Assume a narrow tube packed with very small 2-3 micron particles made of silica.
When we pump solvents such as methanol/water (from the left, inlet) at high flow rates, say 5 mL/min, the back-pressure is very high ~ 500 bar. The right end of the column heats up. One can feel that outlet fluid is quite warm. The outlet of the column is at atmospheric pressure. The inlet to outlet temperature difference can be as high as 10 C. In the literature it is wrongly called frictional heating as if the fluid has warmed up by rubbing against very small silica particles. I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles.
Now, gases can also be used as a mobile phase in the same packed column, e.g. pressurized carbon dioxide. This time the right end of the column is almost freezing, as expected. The pressurized gases are expanding and the gases cool down.
The question is why do compressed liquids warm up when they are allowed to expand, unlike gases? Thanks.
thermodynamics thermal-radiation
thermodynamics thermal-radiation
edited 8 hours ago
M. Farooq
asked 8 hours ago
M. FarooqM. Farooq
1345
1345
$begingroup$
Are you familiar with the property of a fluid called "viscosity?"
$endgroup$
– Chet Miller
6 hours ago
1
$begingroup$
I am glad you didn't start with "Elementary, my dear Watson!"
$endgroup$
– M. Farooq
6 hours ago
$begingroup$
I don’t understand. What are you saying?
$endgroup$
– Chet Miller
6 hours ago
1
$begingroup$
I was intrigued by the way you asked if I were familiar with the fluid property called "viscosity". Of course, everyone is. If your point is viscous heating, this is exactly what that expert was refuting. He said that the heating of the liquid has nothing to do with friction of the liquid with the particles. P.S. "Elementary, my dear Watson" is a Sherlock Holmes' dialogue to his friend Dr. Watson, when he solved a difficult case and wanted to explain it to him.
$endgroup$
– M. Farooq
5 hours ago
add a comment |
$begingroup$
Are you familiar with the property of a fluid called "viscosity?"
$endgroup$
– Chet Miller
6 hours ago
1
$begingroup$
I am glad you didn't start with "Elementary, my dear Watson!"
$endgroup$
– M. Farooq
6 hours ago
$begingroup$
I don’t understand. What are you saying?
$endgroup$
– Chet Miller
6 hours ago
1
$begingroup$
I was intrigued by the way you asked if I were familiar with the fluid property called "viscosity". Of course, everyone is. If your point is viscous heating, this is exactly what that expert was refuting. He said that the heating of the liquid has nothing to do with friction of the liquid with the particles. P.S. "Elementary, my dear Watson" is a Sherlock Holmes' dialogue to his friend Dr. Watson, when he solved a difficult case and wanted to explain it to him.
$endgroup$
– M. Farooq
5 hours ago
$begingroup$
Are you familiar with the property of a fluid called "viscosity?"
$endgroup$
– Chet Miller
6 hours ago
$begingroup$
Are you familiar with the property of a fluid called "viscosity?"
$endgroup$
– Chet Miller
6 hours ago
1
1
$begingroup$
I am glad you didn't start with "Elementary, my dear Watson!"
$endgroup$
– M. Farooq
6 hours ago
$begingroup$
I am glad you didn't start with "Elementary, my dear Watson!"
$endgroup$
– M. Farooq
6 hours ago
$begingroup$
I don’t understand. What are you saying?
$endgroup$
– Chet Miller
6 hours ago
$begingroup$
I don’t understand. What are you saying?
$endgroup$
– Chet Miller
6 hours ago
1
1
$begingroup$
I was intrigued by the way you asked if I were familiar with the fluid property called "viscosity". Of course, everyone is. If your point is viscous heating, this is exactly what that expert was refuting. He said that the heating of the liquid has nothing to do with friction of the liquid with the particles. P.S. "Elementary, my dear Watson" is a Sherlock Holmes' dialogue to his friend Dr. Watson, when he solved a difficult case and wanted to explain it to him.
$endgroup$
– M. Farooq
5 hours ago
$begingroup$
I was intrigued by the way you asked if I were familiar with the fluid property called "viscosity". Of course, everyone is. If your point is viscous heating, this is exactly what that expert was refuting. He said that the heating of the liquid has nothing to do with friction of the liquid with the particles. P.S. "Elementary, my dear Watson" is a Sherlock Holmes' dialogue to his friend Dr. Watson, when he solved a difficult case and wanted to explain it to him.
$endgroup$
– M. Farooq
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.
I don't know who the "expert" was, but he was wrong. The cause of the temperature rise was indeed viscous heating. Viscous "friction" is not the same as dry friction between the liquid with the particles. It is analogous to the heating that occurs in a viscous fluid when it is stirred. The mechanical energy of stirring is dissipated by the viscous behavior of the fluid, and converted to internal energy of the fluid, which translates into a temperature rise. This is exactly how Joule demonstrated the equivalence between mechanical work and increase in internal energy.
In the case of flowing a liquid through a packed column, the mechanical energy loss associated with viscous dissipation in the fluid flow within the pore channels of the packing produces an increase in the internal energy of the exit fluid. This is also a well-known phenomenon in viscous polymer flow through processing equipment, such as transfer lines, spinnerets, and other dies. The rough rule of thumb is a 2 degree temperature rise for every 1000 psi in pressure drop.
In your packed column, application of the open system (control volume) version of the first law of thermodynamics to the flow through the column yields:
$$Delta h = CDelta T+vDelta P=0$$where h is the enthalpy per unit mass of the fluid, C is the specific heat capacity, v is the specific volume and P is the pressure. Viscosity causes the fluid pressure to decrease between the entrance and the exit of the packed bed. The corresponding viscous temperature rise of the fluid is:
$$Delta T=-frac{vDelta P}{C}$$where $Delta P$ is negative.
In the simple case of flow in a tube, the viscous flow equations are observed to accurately predict the amount of viscous dissipation and the corresponding pressure drop. And the above equation is then observed to accurately predict the temperature rise of the fluid.
In the experiments you described, there were two phenomena occurring simultaneously: expansion cooling and viscous heating. In the case of a liquid, the viscous heating wins out over the expansion cooling because a liquid is nearly incompressible, so it can't do any expansion work to cause cooling. In the case of an ideal gas, the expansion cooling exactly cancels the viscous heating, and there is no change in internal energy or temperature. In the case of a real gas, in most cases, the expansion cooling typically wins out (by a little) over the viscous heating, so there is usually a small amount of temperature decrease.
$endgroup$
$begingroup$
Your equation is the same as what others have derived. The Joule experiment is a nice reminder. I did that two decades ago. In ultrahigh pressure liquid chromatography, deltaP is 5000 - 10000 psi. I gather that gentleman must be wrong based on the reasoning aboe. He is a respected figure and we had a chat on this topic. He said the literature is spreading this misconception. It is just adiabatic expansion of a compressed liquid (at the outlet)- any compressed liquid will heat up if allowed to expand. Your thesis is that an expanding liquid would cool, if viscous heating were absent.
$endgroup$
– M. Farooq
3 hours ago
$begingroup$
I agree with your point. "I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.". No problem. This happens in public forums where full profile cannot be shared.
$endgroup$
– M. Farooq
3 hours ago
add a comment |
$begingroup$
I work with ammonia, so I can speak to that.
The liquid gives up its heat as the pressure drops. You may see a slight spike, but the temp will go down as soon as the heat is released.
" I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles." I think you or your expert may have been confused about this term or at least I am confused as you have to put work into the liquid to cause it to heat up. Work in this case is the compression.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.
I don't know who the "expert" was, but he was wrong. The cause of the temperature rise was indeed viscous heating. Viscous "friction" is not the same as dry friction between the liquid with the particles. It is analogous to the heating that occurs in a viscous fluid when it is stirred. The mechanical energy of stirring is dissipated by the viscous behavior of the fluid, and converted to internal energy of the fluid, which translates into a temperature rise. This is exactly how Joule demonstrated the equivalence between mechanical work and increase in internal energy.
In the case of flowing a liquid through a packed column, the mechanical energy loss associated with viscous dissipation in the fluid flow within the pore channels of the packing produces an increase in the internal energy of the exit fluid. This is also a well-known phenomenon in viscous polymer flow through processing equipment, such as transfer lines, spinnerets, and other dies. The rough rule of thumb is a 2 degree temperature rise for every 1000 psi in pressure drop.
In your packed column, application of the open system (control volume) version of the first law of thermodynamics to the flow through the column yields:
$$Delta h = CDelta T+vDelta P=0$$where h is the enthalpy per unit mass of the fluid, C is the specific heat capacity, v is the specific volume and P is the pressure. Viscosity causes the fluid pressure to decrease between the entrance and the exit of the packed bed. The corresponding viscous temperature rise of the fluid is:
$$Delta T=-frac{vDelta P}{C}$$where $Delta P$ is negative.
In the simple case of flow in a tube, the viscous flow equations are observed to accurately predict the amount of viscous dissipation and the corresponding pressure drop. And the above equation is then observed to accurately predict the temperature rise of the fluid.
In the experiments you described, there were two phenomena occurring simultaneously: expansion cooling and viscous heating. In the case of a liquid, the viscous heating wins out over the expansion cooling because a liquid is nearly incompressible, so it can't do any expansion work to cause cooling. In the case of an ideal gas, the expansion cooling exactly cancels the viscous heating, and there is no change in internal energy or temperature. In the case of a real gas, in most cases, the expansion cooling typically wins out (by a little) over the viscous heating, so there is usually a small amount of temperature decrease.
$endgroup$
$begingroup$
Your equation is the same as what others have derived. The Joule experiment is a nice reminder. I did that two decades ago. In ultrahigh pressure liquid chromatography, deltaP is 5000 - 10000 psi. I gather that gentleman must be wrong based on the reasoning aboe. He is a respected figure and we had a chat on this topic. He said the literature is spreading this misconception. It is just adiabatic expansion of a compressed liquid (at the outlet)- any compressed liquid will heat up if allowed to expand. Your thesis is that an expanding liquid would cool, if viscous heating were absent.
$endgroup$
– M. Farooq
3 hours ago
$begingroup$
I agree with your point. "I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.". No problem. This happens in public forums where full profile cannot be shared.
$endgroup$
– M. Farooq
3 hours ago
add a comment |
$begingroup$
I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.
I don't know who the "expert" was, but he was wrong. The cause of the temperature rise was indeed viscous heating. Viscous "friction" is not the same as dry friction between the liquid with the particles. It is analogous to the heating that occurs in a viscous fluid when it is stirred. The mechanical energy of stirring is dissipated by the viscous behavior of the fluid, and converted to internal energy of the fluid, which translates into a temperature rise. This is exactly how Joule demonstrated the equivalence between mechanical work and increase in internal energy.
In the case of flowing a liquid through a packed column, the mechanical energy loss associated with viscous dissipation in the fluid flow within the pore channels of the packing produces an increase in the internal energy of the exit fluid. This is also a well-known phenomenon in viscous polymer flow through processing equipment, such as transfer lines, spinnerets, and other dies. The rough rule of thumb is a 2 degree temperature rise for every 1000 psi in pressure drop.
In your packed column, application of the open system (control volume) version of the first law of thermodynamics to the flow through the column yields:
$$Delta h = CDelta T+vDelta P=0$$where h is the enthalpy per unit mass of the fluid, C is the specific heat capacity, v is the specific volume and P is the pressure. Viscosity causes the fluid pressure to decrease between the entrance and the exit of the packed bed. The corresponding viscous temperature rise of the fluid is:
$$Delta T=-frac{vDelta P}{C}$$where $Delta P$ is negative.
In the simple case of flow in a tube, the viscous flow equations are observed to accurately predict the amount of viscous dissipation and the corresponding pressure drop. And the above equation is then observed to accurately predict the temperature rise of the fluid.
In the experiments you described, there were two phenomena occurring simultaneously: expansion cooling and viscous heating. In the case of a liquid, the viscous heating wins out over the expansion cooling because a liquid is nearly incompressible, so it can't do any expansion work to cause cooling. In the case of an ideal gas, the expansion cooling exactly cancels the viscous heating, and there is no change in internal energy or temperature. In the case of a real gas, in most cases, the expansion cooling typically wins out (by a little) over the viscous heating, so there is usually a small amount of temperature decrease.
$endgroup$
$begingroup$
Your equation is the same as what others have derived. The Joule experiment is a nice reminder. I did that two decades ago. In ultrahigh pressure liquid chromatography, deltaP is 5000 - 10000 psi. I gather that gentleman must be wrong based on the reasoning aboe. He is a respected figure and we had a chat on this topic. He said the literature is spreading this misconception. It is just adiabatic expansion of a compressed liquid (at the outlet)- any compressed liquid will heat up if allowed to expand. Your thesis is that an expanding liquid would cool, if viscous heating were absent.
$endgroup$
– M. Farooq
3 hours ago
$begingroup$
I agree with your point. "I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.". No problem. This happens in public forums where full profile cannot be shared.
$endgroup$
– M. Farooq
3 hours ago
add a comment |
$begingroup$
I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.
I don't know who the "expert" was, but he was wrong. The cause of the temperature rise was indeed viscous heating. Viscous "friction" is not the same as dry friction between the liquid with the particles. It is analogous to the heating that occurs in a viscous fluid when it is stirred. The mechanical energy of stirring is dissipated by the viscous behavior of the fluid, and converted to internal energy of the fluid, which translates into a temperature rise. This is exactly how Joule demonstrated the equivalence between mechanical work and increase in internal energy.
In the case of flowing a liquid through a packed column, the mechanical energy loss associated with viscous dissipation in the fluid flow within the pore channels of the packing produces an increase in the internal energy of the exit fluid. This is also a well-known phenomenon in viscous polymer flow through processing equipment, such as transfer lines, spinnerets, and other dies. The rough rule of thumb is a 2 degree temperature rise for every 1000 psi in pressure drop.
In your packed column, application of the open system (control volume) version of the first law of thermodynamics to the flow through the column yields:
$$Delta h = CDelta T+vDelta P=0$$where h is the enthalpy per unit mass of the fluid, C is the specific heat capacity, v is the specific volume and P is the pressure. Viscosity causes the fluid pressure to decrease between the entrance and the exit of the packed bed. The corresponding viscous temperature rise of the fluid is:
$$Delta T=-frac{vDelta P}{C}$$where $Delta P$ is negative.
In the simple case of flow in a tube, the viscous flow equations are observed to accurately predict the amount of viscous dissipation and the corresponding pressure drop. And the above equation is then observed to accurately predict the temperature rise of the fluid.
In the experiments you described, there were two phenomena occurring simultaneously: expansion cooling and viscous heating. In the case of a liquid, the viscous heating wins out over the expansion cooling because a liquid is nearly incompressible, so it can't do any expansion work to cause cooling. In the case of an ideal gas, the expansion cooling exactly cancels the viscous heating, and there is no change in internal energy or temperature. In the case of a real gas, in most cases, the expansion cooling typically wins out (by a little) over the viscous heating, so there is usually a small amount of temperature decrease.
$endgroup$
I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.
I don't know who the "expert" was, but he was wrong. The cause of the temperature rise was indeed viscous heating. Viscous "friction" is not the same as dry friction between the liquid with the particles. It is analogous to the heating that occurs in a viscous fluid when it is stirred. The mechanical energy of stirring is dissipated by the viscous behavior of the fluid, and converted to internal energy of the fluid, which translates into a temperature rise. This is exactly how Joule demonstrated the equivalence between mechanical work and increase in internal energy.
In the case of flowing a liquid through a packed column, the mechanical energy loss associated with viscous dissipation in the fluid flow within the pore channels of the packing produces an increase in the internal energy of the exit fluid. This is also a well-known phenomenon in viscous polymer flow through processing equipment, such as transfer lines, spinnerets, and other dies. The rough rule of thumb is a 2 degree temperature rise for every 1000 psi in pressure drop.
In your packed column, application of the open system (control volume) version of the first law of thermodynamics to the flow through the column yields:
$$Delta h = CDelta T+vDelta P=0$$where h is the enthalpy per unit mass of the fluid, C is the specific heat capacity, v is the specific volume and P is the pressure. Viscosity causes the fluid pressure to decrease between the entrance and the exit of the packed bed. The corresponding viscous temperature rise of the fluid is:
$$Delta T=-frac{vDelta P}{C}$$where $Delta P$ is negative.
In the simple case of flow in a tube, the viscous flow equations are observed to accurately predict the amount of viscous dissipation and the corresponding pressure drop. And the above equation is then observed to accurately predict the temperature rise of the fluid.
In the experiments you described, there were two phenomena occurring simultaneously: expansion cooling and viscous heating. In the case of a liquid, the viscous heating wins out over the expansion cooling because a liquid is nearly incompressible, so it can't do any expansion work to cause cooling. In the case of an ideal gas, the expansion cooling exactly cancels the viscous heating, and there is no change in internal energy or temperature. In the case of a real gas, in most cases, the expansion cooling typically wins out (by a little) over the viscous heating, so there is usually a small amount of temperature decrease.
edited 4 hours ago
answered 5 hours ago
Chet MillerChet Miller
16.4k2827
16.4k2827
$begingroup$
Your equation is the same as what others have derived. The Joule experiment is a nice reminder. I did that two decades ago. In ultrahigh pressure liquid chromatography, deltaP is 5000 - 10000 psi. I gather that gentleman must be wrong based on the reasoning aboe. He is a respected figure and we had a chat on this topic. He said the literature is spreading this misconception. It is just adiabatic expansion of a compressed liquid (at the outlet)- any compressed liquid will heat up if allowed to expand. Your thesis is that an expanding liquid would cool, if viscous heating were absent.
$endgroup$
– M. Farooq
3 hours ago
$begingroup$
I agree with your point. "I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.". No problem. This happens in public forums where full profile cannot be shared.
$endgroup$
– M. Farooq
3 hours ago
add a comment |
$begingroup$
Your equation is the same as what others have derived. The Joule experiment is a nice reminder. I did that two decades ago. In ultrahigh pressure liquid chromatography, deltaP is 5000 - 10000 psi. I gather that gentleman must be wrong based on the reasoning aboe. He is a respected figure and we had a chat on this topic. He said the literature is spreading this misconception. It is just adiabatic expansion of a compressed liquid (at the outlet)- any compressed liquid will heat up if allowed to expand. Your thesis is that an expanding liquid would cool, if viscous heating were absent.
$endgroup$
– M. Farooq
3 hours ago
$begingroup$
I agree with your point. "I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.". No problem. This happens in public forums where full profile cannot be shared.
$endgroup$
– M. Farooq
3 hours ago
$begingroup$
Your equation is the same as what others have derived. The Joule experiment is a nice reminder. I did that two decades ago. In ultrahigh pressure liquid chromatography, deltaP is 5000 - 10000 psi. I gather that gentleman must be wrong based on the reasoning aboe. He is a respected figure and we had a chat on this topic. He said the literature is spreading this misconception. It is just adiabatic expansion of a compressed liquid (at the outlet)- any compressed liquid will heat up if allowed to expand. Your thesis is that an expanding liquid would cool, if viscous heating were absent.
$endgroup$
– M. Farooq
3 hours ago
$begingroup$
Your equation is the same as what others have derived. The Joule experiment is a nice reminder. I did that two decades ago. In ultrahigh pressure liquid chromatography, deltaP is 5000 - 10000 psi. I gather that gentleman must be wrong based on the reasoning aboe. He is a respected figure and we had a chat on this topic. He said the literature is spreading this misconception. It is just adiabatic expansion of a compressed liquid (at the outlet)- any compressed liquid will heat up if allowed to expand. Your thesis is that an expanding liquid would cool, if viscous heating were absent.
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– M. Farooq
3 hours ago
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I agree with your point. "I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.". No problem. This happens in public forums where full profile cannot be shared.
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– M. Farooq
3 hours ago
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I agree with your point. "I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.". No problem. This happens in public forums where full profile cannot be shared.
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– M. Farooq
3 hours ago
add a comment |
$begingroup$
I work with ammonia, so I can speak to that.
The liquid gives up its heat as the pressure drops. You may see a slight spike, but the temp will go down as soon as the heat is released.
" I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles." I think you or your expert may have been confused about this term or at least I am confused as you have to put work into the liquid to cause it to heat up. Work in this case is the compression.
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add a comment |
$begingroup$
I work with ammonia, so I can speak to that.
The liquid gives up its heat as the pressure drops. You may see a slight spike, but the temp will go down as soon as the heat is released.
" I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles." I think you or your expert may have been confused about this term or at least I am confused as you have to put work into the liquid to cause it to heat up. Work in this case is the compression.
$endgroup$
add a comment |
$begingroup$
I work with ammonia, so I can speak to that.
The liquid gives up its heat as the pressure drops. You may see a slight spike, but the temp will go down as soon as the heat is released.
" I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles." I think you or your expert may have been confused about this term or at least I am confused as you have to put work into the liquid to cause it to heat up. Work in this case is the compression.
$endgroup$
I work with ammonia, so I can speak to that.
The liquid gives up its heat as the pressure drops. You may see a slight spike, but the temp will go down as soon as the heat is released.
" I met a specialist, he said it is the adiabatic expansion of a compressed liquid which leads to heating up and it has nothing to do with friction of the moving liquid with particles." I think you or your expert may have been confused about this term or at least I am confused as you have to put work into the liquid to cause it to heat up. Work in this case is the compression.
answered 6 hours ago
RickRick
695319
695319
add a comment |
add a comment |
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$begingroup$
Are you familiar with the property of a fluid called "viscosity?"
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– Chet Miller
6 hours ago
1
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I am glad you didn't start with "Elementary, my dear Watson!"
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– M. Farooq
6 hours ago
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I don’t understand. What are you saying?
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– Chet Miller
6 hours ago
1
$begingroup$
I was intrigued by the way you asked if I were familiar with the fluid property called "viscosity". Of course, everyone is. If your point is viscous heating, this is exactly what that expert was refuting. He said that the heating of the liquid has nothing to do with friction of the liquid with the particles. P.S. "Elementary, my dear Watson" is a Sherlock Holmes' dialogue to his friend Dr. Watson, when he solved a difficult case and wanted to explain it to him.
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– M. Farooq
5 hours ago